91Ó°ÊÓ

The given atom is Europium- $152$.

Semi empirical binding Energy is given by the expression, $BE=c_{1}A−c_2{A}^{2/3}−c_3\frac{Z(Z−1)}{A^{1/3}}−c_4\frac{{(N−Z)}^2}{A}$.

Where,$c_1=15.8MeV,\text{ }{c}_2=17.8MeV,\text{ }{c}_3=0.71MeV,\text{​â¶Ä‰}{c}_4=23.7MeV$.

$A→$Mass Number,$Z→$atomic Number and$N→$ number of neutrons.

The binding energy calculated by the semi empirical binding energy formula for a europium-$152$atom is note europium has$Z=63$ and$A=152$ .

$\begin{array}{l}N=A−Z\\ N=152−63\\ N=89\end{array}$

Substitute $15.8\text{MeV}$ for $c_1,17.8\text{MeV}$ for $c_3,63$for$Z\text{ },152$ for$A$ for and 89 for $N$ in equation (1).

$\begin{array}{l}BE=c_{1}A−c_2{A}^{2/3}−c_3\frac{Z(Z−1)}{A^{1/3}}−c_4\frac{{(N−Z)}^2}{A}\\ BE=15.8\left(152\right)−17.8{\left(152\right)}^{2/3}−0.71\frac{63(63−1)}{{\left(152\right)}^{1/3}}−23.7\frac{{(89−63)}^2}{152}\\ BE=1269.58\text{MeV}\end{array}$

By the other model, obtain$BE=(Z{M}_H+N{m}_u−M_{Eu})c^2$ .

Rearrange the above equation for$M_{Eu}$.

$M_{Eu}=Z{M}_H+N{m}_u−\frac{BE}{c^2}$

Substitute $1.007825u$ for $M_H,1.008665u$ for $m_u$ and $1269.58\text{MeV}$ for $B.E$ in above equation.

$\begin{array}{c}{M}_{Eu}=\left(63\right)\left(1.007825u\right)+\left(99\right)\left(1.008665u\right)−\frac{\left(1269.58\text{MeV}\right)}{(931.5\text{MeV}/\text{u}u)}\\ M_{Eu}=151.90u\end{array}$